Difference between revisions of "2009 AMC 12A Problems/Problem 9"

Revision as of 20:42, 3 July 2013

Problem

Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$ ?

$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution

Solution 1

As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$ .

To compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$ .

Solution 2

Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$ .

We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$ .

Comparing coefficients we have $a=3$ , $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \boxed{2}$ .

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 16:00, 12 February 2009 (view source) Misof (talk \| contribs) (New page: == Problem == Suppose that <math>f(x+3)=3x^2 + 7x + 4</math> and <math>f(x)=ax^2 + bx + c</math>. What is <math>a+b+c</math>? <math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf...)		Revision as of 20:42, 3 July 2013 (view source) Nathan wailes (talk \| contribs) (→‎See Also) Newer edit →
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	{{AMC12 box\|year=2009\|ab=A\|num-b=8\|num-a=10}}		{{AMC12 box\|year=2009\|ab=A\|num-b=8\|num-a=10}}
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